Let Q be the cube with the set of vertices {(x1, x2, x3) ∈ ℝ³ : x1, x2, x3 ∈ {0, 1}}. Let F be the s

Q: Wait, there are 12 face diagonals and 4 main diagonals. That's 48 pairs! Do I really have to check all of them?

That is the trap! In JEE Advanced, you are rarely asked to perform brute force. Because of the cube's perfect symmetry, any pair of lines will either intersect (distance = 0) or be skew with the exact same geometric relationship. Once you solve for one skew pair, you have solved for all of them. Trust the symmetry!

Q: How do I know if the lines intersect or are skew?

Great question. If you apply the shortest distance formula and the numerator (the scalar triple product) becomes zero, it means the lines are coplanar and intersect. If it is non-zero, they are skew. In this problem, we specifically look for the skew configuration to find the maximum distance.

Q: Can I solve this using pure geometry instead of vectors?

You absolutely can, but it requires a very strong spatial intuition. You would need to find the common perpendicular segment between the two lines. While elegant, it is often prone to visualization errors. The vector method is your 'safety net'—it is algorithmic, robust, and less likely to fail under exam pressure.

Question

Let $Q$ be the cube with the set of vertices $\{(x_1, x_2, x_3) \in \mathbb{R}^3 : x_1, x_2, x_3 \in \{0, 1\}\}$. Let $F$ be the set of all twelve lines containing the diagonals of the six faces of the cube $Q$. Let $S$ be the set of all four lines containing the main diagonals of the cube $Q$; for instance, the line passing through the vertices $(0, 0, 0)$ and $(1, 1, 1)$ is in $S$. For lines $\ell_1$ and $\ell_2$, let $d(\ell_1, \ell_2)$ denote the shortest distance between them. Then the maximum value of $d(\ell_1, \ell_2)$, as $\ell_1$ varies over $F$ and $\ell_2$ varies over $S$, is

Options:

(A) $\frac{1}{\sqrt{6}}$

(B) $\frac{1}{\sqrt{8}}$

(C) $\frac{1}{\sqrt{3}}$

(D) $\frac{1}{\sqrt{12}}$

Sigma Percentile · Accepted Answer

### The Geometry of the Cube: A Journey into 3D Space

Welcome, future engineers! Today, we are not just solving a problem; we are exploring the elegant architecture of a cube. Imagine standing in a room where the walls, floor, and ceiling are perfectly aligned with the $x$, $y$, and $z$ axes.

You are looking at a cube, a shape that defines the very essence of 3D symmetry. We have two types of lines here: the main diagonals, which pierce through the heart of the cube, and the face diagonals, which dance across its surfaces. Our goal is to find the maximum shortest distance between these two families of lines.

### Phase 1: The Power of Symmetry

The first instinct for many students is to panic. There are 12 face diagonals and 4 main diagonals, leading to 48 potential pairs.

However, this is where the JEE Advanced mindset kicks in. A cube is a highly symmetric object. If you rotate it, reflect it, or flip it, the relationship between a main diagonal and a face diagonal remains invariant.

This means that for any pair of lines, they will either intersect (giving a distance of $0$) or they will be skew with the exact same geometric configuration. We only need to solve for one representative pair.

### Phase 2: Setting the Stage

Let us anchor our cube at the origin $O(0,0,0)$. We define our main diagonal, $\ell_2$, as the line connecting the origin to the opposite corner $F(1,1,1)$.

The direction vector for this line is:
$$\vec{b}_2 = (1-0)\hat{i} + (1-0)\hat{j} + (1-0)\hat{k} = \hat{i} + \hat{j} + \hat{k}$$

Now, let us pick a face diagonal, $\ell_1$, that does not intersect our main diagonal. The diagonal on the $xy$-plane connecting $A(1,0,0)$ and $C(0,1,0)$ is perfect. Its direction vector is:
$$\vec{b}_1 = (0-1)\hat{i} + (1-0)\hat{j} + (0-0)\hat{k} = -\hat{i} + \hat{j}$$

### Phase 3: The Vector Toolkit

To find the shortest distance between two skew lines $\vec{r} = \vec{a}_1 + \mu \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \lambda \vec{b}_2$, we use the powerful formula:
$$d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2)|}{|\vec{b}_1 	imes \vec{b}_2|}$$

This formula projects the vector connecting the two lines onto the common normal vector. First, let us find the common normal by taking the cross product $\vec{b}_1 	imes \vec{b}_2$:
$$\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 1 & 0 \ 1 & 1 & 1 \end{vmatrix} = \hat{i}(1) - \hat{j}(-1) + \hat{k}(-2) = \hat{i} + \hat{j} - 2\hat{k}$$

The magnitude of this normal vector is:
$$|\vec{b}_1 	imes \vec{b}_2| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{6}$$

### Phase 4: The Final Calculation

Now, we look at the connecting vector between our chosen points on the lines. We chose the origin for $\ell_2$ and point $A(1,0,0)$ for $\ell_1$. Thus, $\vec{a}_2 - \vec{a}_1 = (0,0,0) - (1,0,0) = -\hat{i}$.

The numerator of our distance formula is the absolute value of the dot product:
$$|(-\hat{i}) \cdot (\hat{i} + \hat{j} - 2\hat{k})| = |-1| = 1$$

Putting it all together, we get:
$$d = \frac{1}{\sqrt{6}}$$

This is the shortest distance for our chosen pair. Because of the symmetry we discussed, this is the maximum possible shortest distance for any pair of lines in the set. You have successfully navigated the 3D space, utilized vector algebra, and leveraged symmetry to bypass brute force. The final answer is **$\frac{1}{\sqrt{6}}$**.

Select Answer:

Visualized Solution

Visualizing the Cube and Lines

Setting Up the Coordinates

Vector Equation of Main Diagonal $ℓ_{2}$

Vector Equation of Face Diagonal $ℓ_{1}$

Shortest Distance Formula

Setting up the Cross Product

Evaluating the Cross Product

Magnitude of the Normal Vector

The Connecting Vector

Computing the Dot Product

Calculating the Shortest Distance

Conclusion and Maximization

The Sigma Insight: Shortest Distance Between Two Skew Lines

The Geometry of the Cube

Phase 1

Phase 2

Phase 3

Phase 4

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